Problem: Let $f(x) = 7x^{2}+x-8$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )?
Solution: The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $7x^{2}+x-8 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = 7, b = 1, c = -8$ $ x = \dfrac{-1 \pm \sqrt{1^{2} - 4 \cdot 7 \cdot -8}}{2 \cdot 7}$ $ x = \dfrac{-1 \pm \sqrt{225}}{14}$ $ x = \dfrac{-1 \pm 15}{14}$ $x =1,-\frac{8}{7}$